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Statistical Learning - Classification (STAT 441/841 CM 763- Fall 2021)

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A solution to a common problem (New)

You may have faced the situation when the math formulas in the body of wikinotes appears extraordinary small (compared to usual font for math formulas). Sometimes this small font helps and sometimes it hurts! One solution to correct this is to simply insert a \, at the beginning of the formula. This will solve the problem without having any effect on the rest of the formula. For example you should write <mth>\,p_{x,y}[/itex] instead of <mth>p_{x,y}[/itex], to see $\,\!p_{x,y}$ instead of $p_{x,y}$.

Examples

Carl Gustav Jung

According to scientists, the Sun is pretty big.<ref>E. Miller, The Sun, (New York: Academic Press, 2005), 23-5.</ref> The Moon, however, is not so big.<ref>R. Smith, "Size of the Moon", Scientific American, 46 (April 1978): 44-6.</ref>

$\sqrt{x^2+2x+1}=|x+1| - \left(\left(\frac{2x^2}{x}\right)^2\right)^2$

Summary During the lecture on May 9th, we have introduced the concepts of pseudo-random variables. We have used the example of “mod” to clarify the basic idea of generating random variable of uniform (0,1). Also, we have used the example of convertible cdf to show how to generate random variables from uniform(0,1). For each of the example in class, the instructor has used Matlab to show how to reach the desired results in Matlab.

Multiplicative Congruential Algorithm We use the operator “mod” e.g. (10 mod 3) = 1

if using the recursive form, (a*x+b mod m) = y Let a=2, b=1, m=3

If x=10 (2*10+1 mod 3) =0 (2*0+1 mod 3) = 1 (2*1+1 mod 3) = 0

Example a=13 b=0 m=31 The first 30 numbers in the sequence are a permutation of integers from 1 to 30 and then the sequence repeats itself.  Values are between 0 and m-1. If the values are normalized by dividing by m-1, then the results is numbers uniformly distributed in the interval [0,1].  There is only a finite number of values—30 in this case.

Question: How to generate exp (lambda) from uniform [0,1]?

Inverse Transform Method

Theorem

Take u~U(0,1), let x=F-1(u)
Then x has distribution function F( ), where F(x)= Pr(X<=x), F-1( ) denotes the inverse function of F( ).


Proof

F(x) = Pr(X<=x)
=Pr (F-1(u)<=x)
=Pr(F(F-1(u))<=F(x))
=Pr(u<=F(x))
=F(x)  (since U~U(0,1))


Example 1

Let f(x)=a*exp^(-a*x)
F(x)=1-exp^(-a*x)
u=1-exp^(-a*x)
x= -1/a*ln(1-u)
F-1(x)= -1/a*ln(1-u)


Therefore, the algorithm is: 1. Draw u~U(0,1) 2. Let x= -1/a*ln(1-u)

Additional Example: Write an algorithm to generate a random variable from F(x)=x^12, 0<x<1 Solution: 1. Generate u~U(0,1) 2. u=x^12

    x=u^(1/12)


3. output x we need to show that Pi si the stationary distribution of this Markov Chain, [pi]=[pi]P detailed balance Remark 1; A common choice for q(y|x) is a normal distribution centered at X with standard deviation b q(y|x)= N (x, b^2) in this case q(y|x) is symmetric.